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Three phase system

Single phase generator

Current induces in the coil as the coil moves in

the magnetic field

Generator for single phase

Note Induction motor cannot start by itself. This

problem is solved by introducing three phase

system

Current produced at terminal

Three phase generator

Instead of using one coil only , three coils are

used arranged in one axis with orientation of

120o each other. The coils are R-R1 , Y-Y1 and

B-B1. The phases are measured in this sequence

R-Y-B. I.e Y lags R by 120o , B lags Y by 120o.

The three winding can be represented by the above

circuit. In this case we have six wires. The emf

are represented by eR , eY, eB.

The circuit can be simplified as follows, where

R1 can be connected to Y and Y1 can be connected

to B. In this case the circuit is reduced to 4

wires.

Since the total emf is zero, R and B1 can be

connected together, thus we arrive with delta

connection system.

Delta connection of three phase windings

Fig. B

Fig.A

Fig. C

- Since the total emf is zero, R and B1 can be

connected together as in Fig.A , thus we arrive

with delta connection system as in Fig. C. - The direction of the emf can be referred to the

emf waveform as in Fig. B where PL is ve (R1-R),

PM is ve (Y-Y1) and PN is ve (B-B1).

Star connection of three phase windings

- R1, Y1 and B1 are connected together.
- As the e.m.f generated are assumed in positive

direction , therefore the current directions are

also considered as flowing in the positive

direction. - The current in the common wire (MN) is equal to

the sum of the generated currents. i.e iRiYiB

. - This arrangement is called four wire

star-connected system. The point N refers to star

point or neutral point.

The instantaneous current in loads L1 , L2 and L3

are

Three-wire star-connected system with balanced

load

For balanced loads, the fourth wire carries no

current , so it can be dispensed

Instantaneous currents waveform for iR, iY and

iB in a balanced three-phase system.

Voltage and current in star connection

- VRY, VYB and VBR are called line voltage
- VR, VY and VB are called phase voltage

From Kirchoff voltage law we have

In phasor diagram

For balanced load VR , VY and VB are equaled but

out of phase

VRY VL?30? VYB VL?-90? VBR VL?150?

VR VP?30? VY VP?-90? VB VP?150?

therefore

then

and

Voltage and current in Delta connection

- IR, IY and IB are called line current
- I1, I2 and I3 are called phase current

From Kirchoff current law we have

VP

VL

In phasor diagram

Since the loads are balanced, the magnitude of

currents are equaled but 120o out of phase. i.e

I1 I2I3 ,IP Therefore-

I1 VP?30? I2 VP?-90? I3 VP?150?

IR IL?30? IY IL?-90? IB IL?150?

Where IP is a phase current and IL is a line

current

Thus IRIYIB IL

Hence

Unbalanced load

Example 1

In a three-phase four-wire system the line

voltage is 400V and non-inductive loads of 5 kW,

8 kW and 10 kW are connected between the three

conductors and the neutral. Calculate (a) the

current in each phase (b) the current in the

neutral conductor.

Voltage to neutral

Current in 10kW resistor

Current in 8kW resistor

Current in 5kW resistor

Resolve the current components into horizontal

and vertical components.

Example 2

- A delta connected load is arranged as in Figure

below. The supply voltage is 400V at 50Hz.

Calculate - The phase currents
- The line currents.

(a)

I1 is in phase with VRY since there is only

resistor in the branch

In branch between YB , there are two components ,

R2 and X2

In the branch RB , only capacitor in it , so the

XC is -90 out of phase.

(b)

q30o

q 71o 34 -60o 11o 34

q 180-30o-11o 34 138o 34

Power in three phase

Active power per phase IPVP x power factor

Total active power 3VPIP x power factor

If IL and VL are rms values for line current and

line voltage respectively. Then for delta (?)

connection VP VL and IP IL/?3. therefore

For star connection (?) VP VL/?3 and IP

IL. therefore

Example 3

- A three-phase motor operating off a 400V system

is developing 20kW at an efficiency of 0.87 p.u

and a power factor of 0.82. Calculate - The line current
- The phase current if the windings are

delta-connected.

(a) Since

And line current IL40.0A

(b) For a delta-connected winding

Example 4

Three identical coils, each having a resistance

of 20? and an inductance of 0.5 H connected in

(a) star and (b) delta to a three phase supply of

400 V 50 Hz. Calculate the current and the

total power absorbed by both method of

connections.

First of all calculating the impedance of the

coils

where

Star-connection

Since it is a balanced load

Power absorbed

Star connection

Example 5

A balanced three phase load connected in star,

each phase consists of resistance of 100 ?

paralleled with a capacitance of 31.8 ?F. The

load is connected to a three phase supply of 415

V 50 Hz. Calculate (a) the line

current (b) the power absorbed (c) total

kVA (d) power factor .

Admittance of the load

where

Line current

Volt-ampere per phase

Active power per phase

Total active power

(b)

Reactive power per phase

Total reactive power

Total volt-ampere

(c)

(d)

Power Factor cos? cos 45? 0.707 (leading)

Example 6

A three phase star-connected system having a

phase voltage of 230V and loads consist of non

reactive resistance of 4 ?, 5 ? and 6?

respectively. Calculate (a) the current in each

phase conductor (b) the current in neutral

conductor and (c) total power absorbed.

(b)

X-component 46 cos 30? 38.3 cos 30? - 57.5

15.5 A Y-component 46 sin 30? - 38.3 sin 30?

3.9 A

Therefore

(c)